Element of Surface Area

In order to find an element of area  
\[d \mathbf{S}\]
  surface  
\[S\]
, it is necessary to find vectors  
\[\vec{u}. \; \vec{v}\]
  in the surface then we can define an element of surface area as  
\[d \mathbf{S}= d \vec{u} \times d \vec{v}\]
.
Because  
\[d \mathbf{S}\]
  is defined as the cross product of two vectors, it is itself a vector and is perpendicular to both  
\[d vec{u}\]
  and  
\[d \vec{v}\]
.
Suppose the equation of a surface is  
\[z=a^2-x^2-y^2\]
. Write this as  
\[a^2=x^2+y^2+z\]
.
Then  
\[0=2xdx+2ydy+dz\]
. Along the line where  
\[y=0\]
  this becomes  
\[0=2xdx+dz\]
  corresponding to the vector  
\[ \begin{pmatrix}dx\\0\\-2xdx\end{pmatrix} = \begin{pmatrix}1\\0\\-2x\end{pmatrix} dx \]
.
Along the line where  
\[x=0\]
  this becomes  
\[0=2ydy+dz\]
  corresponding to the vector  
\[ \begin{pmatrix}0\\dy\\-2ydy\end{pmatrix} = \begin{pmatrix}0\\1\\-2y\end{pmatrix} dy \]
.
The cross product of these two vectors is  
\[ \begin{pmatrix}1\\0\\-2x\end{pmatrix} dx \times \begin{pmatrix}0\\1\\-2y\end{pmatrix} dy = \begin{pmatrix}2x\\2y\\1\end{pmatrix} dxdy \]
.
This element of area is a vector perpendicular to the surface at the point  
\[(x,y,a^2-x^2-y^2\]
.

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